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2 years ago

10

Demonstrate an understanding of Stoichiometry by describing and calculating the components required for a stoichiometric evaluat

ion and performing stoichiometric evaluations for determining the limiting reactant and percent yield.

1 answer:

NeTakaya2 years ago

8 0

Explanation:

Iam sorry I don't know but why Iam messaging iss because when more people message it usually appears to more people so someone else will be able to help you:)

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If a tractor trailer and a car are both traveling 50km/h when they hit which will have the greatest change in speed and directio

den301095 [7]

The tractor trailer will push the car when they hit so therefore the car will have to greatest change

4 0

2 years ago

How many significant figures are in 3.02 x 108 ?<br> a<br> 2<br> b 4<br> C 5<br> d 3

bija089 [108]

Answer:

C

Explanation:

the answer has 5 Numbers

7 0

2 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)

vichka [17]

Answer:

Kb → 1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m = Kb → 1.56 °C / m

3 0

3 years ago

The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

$2N_2O \rightarrow 2N_2 + O_2$

rate constant $k = 1.94\times 10^{-4} min^{-1}$

Half life $t_{1/2} = \frac{0.693}{k} = 3572 min$

Initial pressure $N_2 O = 4.70 atm$

Pressure after 3572 min = P

According to first order kinematics

$k = \frac{1}{t} ln\frac{4.70}{P}$

$1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}$

solving for P we get

P = 2.35 atm

$2N_2O \rightarrow 2N_2 + O_2$

initial 4.70 0 0

change -2x +2x +x

final 4.70 -2x 2x x

pressure of $O_2$ after first half life = 2.35 = 4.70 - 2x

x = 1.175

pressure of $N_2$ after first half life = 2x = 2(1.175) = 2.35 ATM

Total pressure = 2.35 + 2.35 + 1.175

= 5.875 atm

5 0

3 years ago

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